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EMT
1. Two homogeneous isotropic dielectrics
are separated by plane z = 0.
For z > 0 ; ϵr = 4 and E 3ax – 4ay 6az= +
kV/m
If ϵr = 3 for z < 0 then find,
A. E for z < 0
B. The angles that electric field makes
with the normal to the plane z = 0 for z
> 0 and for z < 0
C. The energy densities in both the
dielectrics
D. The energy within a cube of side of 4
m and centred at (1, –3, 5)
Sol.
A) Let x y zr 114 and E 3a – 4a 6a = = +
r23 =
Since mediums are separated by z = 0
plane za will be normal to the plane
zn1E 6a =
t 1 n1 1&E E –E=
x yt1E 3a – 4a =
Now t t2 1E E=
x yt2E 3a – 4a = …(i)
n n2 1and D D=
r n r n2 2 1 1E E =
zn2
4E · 6 a
3 =
zn2E 8a = …(ii)
Adding (i) and (ii)
2 t n1 1E E E= +
x y z2E 3a – 4a 8a = + kV/m For z<0
B)
Now
zn1E 6a=
n1E 6 =
x yt1and E 3a – 4a=
2 2t1
E 3 4 = +
t1E 5 =
t11
n1
EtanQ
E =
5
6=
∴ Q1 = 39.800
∴∝1 = 90 – Q1
= 90 – 39.80
∴∝1 = 50.19°
So E1 makes 50.19° angle with the
normal to z = 0 plane
zn2E 8a=
n2E 8 =
x yt2and E 3a – 4a=
2 2t2
E 3 4 = +
t2E 5 =
t22
n2
EtanQ
E =
5
8=
∴ Q2 = 32°
∴∝2 = 90 – Q2
= 90 – 32
∴∝2 = 58°
So E2 makes 58° angle with the normal
to z = 0 plane.
C) The energy density is given by,
21 1 1
1WE E
2=
( )2 2 2 69
1 14 3 4 6 10
2 36 10= + +
31WE 1.07 mJ / m =
∴ Energy density for z > 0 will be 1.07
mJ/m3
22 2 2
1now WE E
2=
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( )2 2 2 69
1 13 3 4 8 10
2 36 10= + +
∴ WE2 = 1.4 mJ/m3
∴ Energy density for z < 0 will be 1.4
mJ/m3
D) Given cube has centre (1, –3, 5) and
it has side of 4 m so it completely lies
for z > 0
∴ the cube will have energy density 1.07
mJ/m3
As it has side of 4 m
It will cover the region
–1 < x < 3 ; –5 < y < –1 ; 3 < z < 7
2 2WE WE dV =
3 –1 7
2
x –1 y –5 z 3
WE dx dy dz= = =
=
= 1.07 × 4 × 4 × 4 × 10–3
∴ WE = 69.03 mJ
2. The magnetic field of a plane was
travelling in free space is given by,
( )( )= y xsH 0.05 cos t – j6 z a – ja A/m
then find
A. Frequency and direction of
propagation
B. Corresponding electric field vector
C. polarization of the wave
D. time average power density
associated with wave
Sol.
A) Hs = 0.05 cos(ωt – j6πz) ( )y xa – ja
A/m
∴ the wave is travelling along +ve z
direction and β = 6π
pnow V
=
As the wave is travelling through free
space
8pV 3 10
= =
∴ ω = 6π × 3 × 108
∴ f = 9 × 108
∴ Frequency = 900 MHz
B) Now Hs=0.05cos(ωt – j6πz)( y xa – ja )
( )
( )
=
ys
x
H 0.05 cos t – j6 z a
– 0.05jcos t – j6 z a
( ) y1let H 0.05 cos t – j6 z a=
( ) x2and H –0.05jcos t – j6 z a=
Enow 120 and E H P
H= =
1E 0.05 120 18.8 = =
∴ E1 direction = H1 direction ×
propagation direction
= y za a
x1E direction a =
( ) x1E 18.8 cos t – j6 z a =
Similarly
2E 0.05 120 18.8= =
And E2 direction = H2 direction ×
propagation direction ( ) ( )x z– j a a=
y2E direction ja =
Now E = E1 + E2
( )( )x yE 18.8 cos t – j6 z a ja V /m = +
C) ( )( )x yE 18.8 cos t – j6 z a ja V /m= +
We calculate polarization at z = 0
( ) ( )( )x yE 0,t 18.8cos t a ja = +
( ) x yE 0,t 18.8cos t a –18.8sin t a =
As E1 = E2 it will be circular polarization
let’s draw E vector at different time
intervals.
At t = 0 xE 18.8 a=
At t = T/8
=
x y
2 T 2 TE 18.8cos a –18.8in a
T 8 T 8
x y1 1
E 18.8 a – a2 2
=
At t = T/4
=
x y
2 T 2 TE 18.8cos a –18.8sin a
T 4 T 4
yE –18.8a=
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∴It is left circular polarization.
D) Time average power density
associated with wave is
( )= *s s
1P E H
2
( ) ( )x y y x1
18.8 a ja 0.05 a ja2
= + +
( )z z0.94
a a2
= +
2zP 0.94a W / m=
3.
For the circuit as shown above calculate
A. Reflection co-efficient and VSWR
B. Input impedance seen by transmitter
C. The time averaged power delivered to
the transmission line by the transmitter
D. The time averaged power delivered to
antenna
E. Total power supplied by the
transmitter
Sol.
A)Reflection co-efficient (Γ)
( )
( ) = =
+ +
L o
L o
200 – j50 – 50Z – Z
Z Z 200 – j50 50
= 0.6154 – j0.0769
∴ Γ = 0.6202 < –0.1244
0.6202 =
+ =
1VSWR
1–
1 0.6202
1 – 0.6202
+=
∴ VSWR = 4.27
B) pNow V
=
pV
=
p
2 f
V
=
8
8
2 5 10 2.1
3 10
=
15.18 rad /m=
Now
L oin o
o L
Z cos jZ sinZ Z ·
Z cos jZ sin
+ =
+
( ) ( ) ( )
( ) ( ) ( )
+ =
+
200 – j50 cos 15.18 4 j50sin 15.18 450 ·
50cos 15.18 4 200 – j50 sin 15.18 4
inZ 14.664 – j24.17=
C) Equivalent circuit can be drawn as,
inin s
in G
ZV V ·
Z R =
+
14.664 – j24.1760
14.664 – j24.17 60=
+
inV 16.3574 – j14.127V =
∴ Vin = 21.613 < –40.81 V
=inV 21.613 V
= inin
in
Vnow I
Z
21.613 –40.81
14.664 – j27.14
=
= +inI 0.655 j0.298
∴ Iin = 0.7 < 20.8
=inI 0.7 A
Now power will be dissipated only across
resistive part of the load
( ) ( )= 2
in in1
I Real Z2
( )2
in1
P 0.7 14.6642
=
∴ Pin = 3.592 watt
D) As the transmission line is lossless,
total power delivered to the transmission
line will be completely delivered to the
antenna. The time average power
delivered to the antenna is 3.592 watt.
E) Power dissipated in internal
impedance RG is
( )= 2
G in G1
P I R2
( )21
0.7 602
=
∴ PG = 14.7 watt
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F) Total power supplied by the
transmitter will be,
= +T in GP P P
= 3.592 + 14.7
=TP 18.292 watt
4. A. Define dominant mode in a
rectangular waveguide and find the
power carried by the dominant mode for
a > b by using transverse components.
B. Plot the TE10 mode and how to extract
the TE10 mode effectively.
Sol.
A) Dominant mode: The mode which is
having highest cut off wave length (or)
lowest cut off frequency.
2 2
2
( ) ( )C
ab
mb na =
+
Lowest TM mode is TE11
11 112 2
01
10
2
2
2
C C
C
C
abTM TE
b a
TE b
TE a
= =+
=
=
As a > b i.e., TE10 is the dominant mode
Field components for TE10
cos cosZ
m nH C x y
a b
=
2
2
2
2
cos cos
sin cos
= −
=
=
− =
ZX
Zy
HjE
yh
j n m nC x y
b a bh
HjE
xh
j m m nC x y
a a bh
2
2
2
2
sin cos
cos sin
= −
=
= −
=
ZX
Zy
HH
xh
m m nC x y
a a bh
HH
yh
n m nC x y
b a bh
For, 10 cos
0
=
= =
z
x y
TE H C xa
and E H
2
2
sin
sin
y
x
jE C x
a ah
jH C x
a ah
− =
=
Power carried by waveguide
.
s
p p dS=
1Re[ *]2
p E H=
2 2
22 2
2
2 2
1ˆ ˆRe sin sin
2
( )1
ˆsin2 ( )
− − =
=
j jC x y x C xa a a ah h
Ca
p x zah
22max ˆsin
2 TE
Ep x z
Z a
=
22max
0
. sin2
a
z zTEs x
EP p dS b x dx
Z a
=
= =
2max
2max
10
2 2
4
TE
TETE
E ab
Z
EP ab
Z
=
=
B) We have
sinyE xa
As the maximum E is existing at
midpoint of the broader dimension, it is
effective if making use of probe or E-
coupling.
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6
5. A. Find the radiation efficiency of an
Hertzian dipole metal wire having radius
1.8 mm, length 2 meter and σ = 6 × 109
s/m and it is operating at frequency of 2
MHz.
B. An end-fire array consists of λ/2
radiations with axes at right angles to
the line of the array required to have a
power gain of 30. Determine the array
length and the width of a major lobe
between the nulls.
Sol.
A) For Hertzian dipole, Radiation
resistance (Rr) 2
2r
dlR 80
=
22 dl f
80c
=
262
8
2 2 1080
3 10
=
Rr = 0.14 Ω
And loss resistance (RL)
=
L
dl fµR
2 a
Now dl = 2m, a = 1.8 × 10–3,
µ = 4π × 10–7
f = 2 × 106, σ = 6 × 109
=
6 –7
L –3 9
2 2 10 4 10R
2 1.8 10 6 10
∴ RL = 6.41 × 10–3 Ω
∴ RL = 0.00641 Ω
∴ Radiation efficiency (η)
r
r L
R
R R=
+
0.14
0.14 0.00641=
+
∴ η = 0.9561
∴ Radiation efficiency = 95.61%
B) For an end fire array
=
LD 4
=
L30 4
∴ L = 7.5λ
∴ Array length = 7.5λ
Now width of a major lobe between the
nulls is given as,
2BWFN 114.6
L /=
2114.6
7.5=
∴ BWFN = 59.18°
∴ Width of major lobe between the nulls
is 59.18°.
***